UPSC EPFO Exam Date has been released. The exam will be conducted on 2nd July 2023. A total of 418 vacancies were announced for the post of
Enforcement Officer/Accounts Officer and 159 vacancies for APFC. The selection process will consist of two stages: a recruitment test & an interview round. The weightage for the recruitment chạy thử will be 75%, and the weightage for the interview will be 25%. Candidates with a bachelor's degree và under the age of 35 will be considered for the recruitment drive. Khổng lồ enhance your exam preparation, refer to lớn the UPSC EPFO Previous Year Papers. Also, attempt UPSC EPFO demo Series.

Bạn đang xem: What is the formula of x3+y3+ z3


Win over the concepts of Algebra và get a step ahead with the preparations for Quantitative Aptitude with Testbook.
*

*

*

*

344 Total Tests with
*
5 không tính tiền Tests
Exam Date
Eligibility Criteria
Salary and Job Profile
Selection Process
Syllabus
Result
Admit Card
Answer Key
Cut Off
Previous Year Papers
Books
Exam Analysis
Exam Centres
Current Affairs
Interview
General English
Preparation Strategy
Notes
Enforcement Officer
General Accounting Principles
Social Security in India
UPSC EPFO vs SSC CGLAccounts Officer
APFCLabour Laws
Difference Between UPSC APFC vs EO/AO
Q1. Five years ago the sum of the ages of A & B was 58 years. The difference between B's age 8 years ago and A's age 8 years hence is 16 years. The ratio of the present ages of A & B is:
Q2. What approximate will come in the place of the question mark ‘?’ in the following question?5.78% of 799.94 + ?% of 9.67 = 10.94 × 2.99 + √100 × 2.98
Q3. What approximate value will come in the place of the question mark ‘?’ in the following question?9.99% of 29.906 + 299.84% of 54.908 – 49.86% of 149.59 = ?
Q4. What approximate value should come in place of the question mark (?) in the following question?63.92 ÷ √(255.89) × 24.91% of (2.99)3 = (?)3
Q5. What approximate value should come in the place of the question mark ‘?’ in the following question?728.821/3+√1155.98 + 6.142÷ 2.992+ 1.970=√?
Q6. What approximate value should come in the place of the question mark ‘?’ in the following question?11.11% of 99.17+ 22.22% of 98.87 -9.89% of 100.12 = ?
Q7. In a group of 160 people, 100 people like tennis & 20 people like both cricket & tennis. If each one likes at least one of these two games,then find the ratio between the number of people who like only cricket and the number of people who lượt thích only tennis.
Q8. A certain sum of money is distributed among Ravi, Rahul, và Raj in ratio 8 : 5 : 7 in such a way that giới thiệu of Ravi was Rs. 1000 less than that the sum of mô tả of Rahul và Raj. Find the difference between the shares of Ravi and Raj?
Q9. Radha's expenditure is 20% of her income. If the income for next month is increased by 20%, and the amount of savings remains the same, then find the percentage increase in expenditure of Radha.
Q10. A can vị a piece of work in 22 days while B alone can complete the whole work in 26 days. They work together for 5 days then A left the work và remaining work can done by B alone. In how many days will finish the whole work?
Daily Live Classes250+ kiểm tra series
Study Material và PDFQuizzes With Detailed Analytics+ More Benefits
I want to lớn factorize the polynomial $x^3+y^3+z^3-3xyz$. Using chiaseyhoc.comematica I find that it equals $(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$. But how can I factorize it by hand?


*

Note that (can be easily seen with rule of Sarrus)$$ eginvmatrix x & y & z \ z và x và y \ y & z và x \ endvmatrix=x^3+y^3+z^3-3xyz$$

On the other hand, it is equal to (if we add to the first row 2 other rows)$$ eginvmatrix x+y+z và x+y+z & x+y+z \ z và x và y \ y & z và x \ endvmatrix=(x+y+z)eginvmatrix 1 và 1 & 1 \ z và x và y \ y & z và x \ endvmatrix=(x+y+z)(x^2+y^2+z^2-xy-xz-yz)$$ just as we wanted. The last equality follows from the expansion of the determinant by first row.

Xem thêm: Khi nói về quá trình hình thành loài mới bằng con đường cách li địa lí phát biểu nào sau đây đúng


nói qua
Cite
Follow
answered Dec 27, 2013 at 15:17
*

Elensil
Elensil
1,08388 silver badges1313 bronze badges
$endgroup$
1
địa chỉ cửa hàng a bình luận |
20
$egingroup$
eginalignx^3+y^3+z^3-3xyz\&= x^3+y^3+3x^2y+3xy^2+z^3-3xyz-3x^2y-3xy^2\&= (x+y)^3+z^3-3xy(x+y+z)\&= (x+y+z)((x+y)^2+z^2-(x+y)z)-3xy(x+y+z)\&= (x+y+z)(x^2+2xy+y^2+z^2-yz-xz-3xy)\&= (x+y+z)(x^2+y^2+z^2-xy-yz-zx)endalign


chia sẻ
Cite
Follow
edited Aug 23, 2021 at 21:40
*

Someone
13477 bronze badges
answered Oct 29, 2013 at 10:23
*

Yurii Savchuk
Yurii Savchuk
3,7511515 silver badges2222 bronze badges
$endgroup$
địa chỉ cửa hàng a comment |
19
$egingroup$
Consider the polynomial $$(lambda - x)(lambda - y)(lambda - z) = lambda^3 - alambda^2+blambda-c ag*1$$We know$$egincasesa = x + y +z\ b = xy + yz + xz \ c = x y zendcases$$ Substitute $x, y, z$ for $lambda$ in $(*1)$ and sum, we get$$x^3 + y^3 + z^3 - a(x^2+y^2+z^2) + b(x+y+z) - 3c = 0$$This is equivalent to$$eginalign x^3+y^3+z^3 - 3xyz= và x^3+y^3+z^3 - 3c\= và a(x^2+y^2+z^2) - b(x+y+z)\= & (x+y+z)(x^2+y^2+z^2 -xy - yz -zx)endalign$$


cốt truyện
Cite
Follow
answered Oct 29, 2013 at 11:16
*

achille huiachille hui
120k66 gold badges178178 silver badges335335 bronze badges
$endgroup$
2
showroom a comment |
12
$egingroup$
Use Newton"s identities:

$p_3=e_1 p_2 - e_2 p_1 + 3e_3$ & so $p_3-3e_3 =e_1 p_2 - e_2 p_1 = p_1(p_2-e_2)$ as required.

Here

$p_1= x+y+z = e_1$

$p_2= x^2+y^2+z^2$

$p_3= x^3+y^3+z^3$

$e_2 = xy + xz + yz$

$e_3 = xyz$


chia sẻ
Cite
Follow
edited Oct 29, 2013 at 11:31
achille hui
120k66 gold badges178178 silver badges335335 bronze badges
answered Oct 29, 2013 at 10:29
lhflhf
212k1515 gold badges227227 silver badges538538 bronze badges
$endgroup$
add a phản hồi |
11
$egingroup$
A polynomial from $chiaseyhoc.combbQ$ is a polynomial from $chiaseyhoc.combbQ$, so it can be viewed as a polynomial in $z$ with coefficients from the integral domain $chiaseyhoc.combbQ$.$$p(z)=z^3-3xy cdot z +x^3+y^3$$

So we can try our methods to factor a polynomial of degree 3 over an integral domain:If it can be factored then there is a factor of degree $1$, we call it $z-u(x,y)$ and $u(x,y)$ divides the constant term of $p(z)$ which is $x^3+y^3$. The latter is can be factored khổng lồ $(x+y)(x^2-xy+y^2)$ We kiểm tra each of the possible values $(x+y), -(x+y), (x^2-xy+y^2), -(x^2-xy+y^2)$ for $u(x,y)$ & find that only $p(-x-y)=0$. So $z-(-x-y)$ is a factor.

Note:

One can use Kronecker"s method

to reduce the factorization of a polynomial of $chiaseyhoc.combbQ$ lớn factoring polynomials in $chiaseyhoc.combbQ$, khổng lồ reduce the factorization of polynomial of $chiaseyhoc.combbQ$ khổng lồ factoring polynomials in $chiaseyhoc.combbQ$to reduce the factorization of polynomial of $chiaseyhoc.combbQ$ to factoring numbers in $chiaseyhoc.combbZ$

This factoring is possible in a finite number of steps but the number of steps may become lớn high for practical purpose.

An integral domain name is a commutative ring with $1$, where the following holds:$$a e 0 land b e 0 implies ab e 0$$For polynomials $f$, $g$, $h$ $in I$ this guarantees:$$f=g cdot h implies extdegree(f)= extdegree(g) + extdegree(h) ag1$$compare this to lớn $chiaseyhoc.combbZ_4$ which is no integral domain and $(2z^2+1)^2 equiv 1$ & so $(2z^n+1) mid 1$. So the polynomial $1$ of degree $0$ has infinitely many divisor.If $I$ is an integral domain $(1)$ guarantees that $z^3+az^2+bz+c in I$ has a linear factor và therefore zero in $I$ if it is not irreduzible.